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(B) Let the roots of the equation $x^{2} + (2 - \lambda)x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta =

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$2\sqrt{5}$

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(B) Let the roots of the equation $x^{2} + (2 - \lambda)x + (10 - \lambda) = 0$ be $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta = \lambda - 2$ and $\alpha\beta = 10 - \lambda$.The sum of the cubes of the roots is $S = \alpha^{3} + \beta^{3} = (\alpha + \beta)^{3} - 3\alpha\beta(\alpha + \beta)$.Substituting the values,$S = (\lambda - 2)^{3} - 3(10 - \lambda)(\lambda - 2)$.$S = (\lambda - 2)[(\lambda - 2)^{2} - 3(10 - \lambda)] = (\lambda - 2)(\lambda^{2} - 4\lambda + 4 - 30 + 3\lambda) = (\lambda - 2)(\lambda^{2} - \lambda - 26)$.$S = \lambda^{3} - \lambda^{2} - 26\lambda - 2\lambda^{2} + 2\lambda + 52 = \lambda^{3} - 3\lambda^{2} - 24\lambda + 52$.To find the minimum,set $\frac{dS}{d\lambda} = 3\lambda^{2} - 6\lambda - 24 = 0$.$3(\lambda^{2} - 2\lambda - 8) = 0 \implies 3(\lambda - 4)(\lambda + 2) = 0$.For minimum,$\frac{d^{2}S}{d\lambda^{2}} = 6\lambda - 6$. At $\lambda = 4$,$6(4) - 6 = 18 > 0$,so it is a minimum.The difference of the roots is $|\alpha - \beta| = \sqrt{(\alpha + \beta)^{2} - 4\alpha\beta} = \sqrt{(\lambda - 2)^{2} - 4(10 - \lambda)}$.For $\lambda = 4$,$|\alpha - \beta| = \sqrt{(4 - 2)^{2} - 4(10 - 4)} = \sqrt{4 - 24} = \sqrt{-20}$.The magnitude of the difference is $|\sqrt{-20}| = \sqrt{20} = 2\sqrt{5}$.

If $\lambda \in R$ is such that the sum of the cubes of the roots of the equation $x^{2} + (2 - \lambda)x + (10 - \lambda) = 0$ is minimum,then the magnitude of the difference of the roots of this equation is

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